Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Could someone please clarify whether my calculation on the following limit problem is correct?

Determine the following limit:

$\lim_{x \to \frac{\pi}{2}} \frac{\sin^2x-1}{\sin x-1}$

$\lim_{x \to \frac{\pi}{2}} \frac{\sin^2x-1}{\sin x-1}$ = $\lim_{x \to \frac{\pi}{2}} \sin x\frac{\sin x-1}{\sin x-1} = \lim_{x \to \frac{\pi}{2}}\sin x \frac{1}{1} = \lim_{x \to \frac{\pi}{2}} \sin x = 1$

Thank you.

share|cite|improve this question
2  
$\sin^2{x}-1\not= \sin{x}(\sin{x}-1)$. Instead, it should be $(\sin{x}-1)(\sin{x}+1)$ – Golbez Aug 23 '14 at 14:04
up vote 7 down vote accepted

$$\lim \frac{\sin^2 x-1}{\sin x-1}=\lim_{x \to \frac{\pi}{2}} \frac{(\sin x-1) \cdot (\sin x+1)}{\sin x-1}=\lim_{x \to \frac{\pi}{2}} (\sin x+1)=2$$

share|cite|improve this answer

Using $a^2-b^2=(a+b)(a-b)$, $$\frac{\sin^2 (x) -1}{\sin (x)-1}=\sin (x)+1$$ The rest is easy.

share|cite|improve this answer
    
Info on this link will help you in formating your replies so that they are legible meta.math.stackexchange.com/questions/5020/… – Vikram Aug 23 '14 at 14:14

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.